# Escape Velocity

1. Nov 16, 2004

### cougar_21

Question:

The radius of a neutron star is 750 times smaller than the Earth's radius, and its mass is 1.8 times larger than the Earth's mass. What is the escape velocity from the surface of a neutron star? (Ignore the fact that, at high speeds, one should not really use mv^2/2 for the kinetic energy.)

If my calculations are correct:
Rn=-4.7711x10^9 m
Mn= 1.0757x10^30 kg

I know that E= K + U in whick K is mv^2/2 and U = GmM/R but don't know how to advance, please help

Does vesc= square root of (2GMn/Rn) ?

Last edited: Nov 16, 2004
2. Nov 16, 2004

### Andrew Mason

You have omitted the minus sign from potential energy. The condition for escape is K + U > 0 ($U \rightarrow 0$ as $R \rightarrow \infty$).

AM

3. Nov 16, 2004

### gnome

First, as Andrew said, U = -GmM/R. It is defined that way so that U = 0 when the distance between
the objects is "infinite".

Assuming your object is given some initial velocity and thereafter no force acts on it
other than gravity from the star, the total energy is constant, and at any given moment
E = K + U = (mv^2)/2 - GmM/R

Escape velocity means just enough to get to such a distance from the star that it's gravity
can no longer affect the object; i.e. theoretically the distance approaches
infinity, all of the initial kinetic energy is converted to potential energy and
therefore the final velocity = 0.

Holding total energy constant,
(1/2)mvi2 - GmM/Ri = (1/2)mvf2 - GmM/Rf
but since vf = 0 this becomes
(1/2)mvi2 - GmM/Ri = - GmM/Rf
rearranging terms:
(1/2)mvi2 = GmM/Ri - GmM/Rf
vi2 = 2GM(1/Ri - 1/Rf)
and since Rf -> ∞ the second fraction approaches 0 so this gives your equation:
vi2 = 2GM/Ri

4. Nov 17, 2004

### cougar_21

thank you andrew and gnome, i would get the square of a negative answer and knew that was not right, but indeed I missed the minus sign. Thanks for the explanation gnome it helped me understand what i was doing !! Thank you both bunches XD!!